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Proof step 232 from proof inv-truncate (previous step)

# buf-opt.buf = VFS_PAGE_SIZE, fs0[ri0[inodekey(ino)]].nlink = 0 ↔ inodekey(ino) ∈ ro, m0 = 0 ∧ m ≠ ROOT_INO ↔ inodekey(m) ∈ ro, VFS_PAGE_SIZE = n0 + m3 % VFS_PAGE_SIZE, fs0[ri0[inodekey(m)]] = inodenode(key, md, false, m0, m1, m2), ro.inodes?, fo.inodes?, buf-opt.some?, file-node-cons(fs0, ri0), log-cons(log0, fs0), orphans-cons(fs0, ri0, ro), replay-cons-strong(fs0, ri0, ro, fi, fo, log0), ri-inj(ri0), ri-inj(fi), ubifs-cons(fs0, ri0, ro, fi, fo, log0), ri0[inodekey(m)] ∈ fs0, inodekey(m) ∈ ri0, inodekey(ino) ∈ ri0, m3 ≤ n1 ⊦ m3 % VFS_PAGE_SIZE = 0, (adr2 = ri0[inodekey(ino)] ⊃ inodenode(inodekey(m), md0, false, m0, m1, n1) ;(adr1 = ri0[inodekey(ino)] ⊃ truncnode(inodekey(m), m3);fs0[ri0[inodekey(ino)]]) ) .nlink = 0 ↔ inodekey(ino) ∈ ro, adr1 = adr3, adr1 = adr2, adr2 = adr3, adr2 = ri0[inodekey(ino)], adr3 = ri0[inodekey(ino)], ino = ROOT_INO, m = ino, adr1 ∈ fs0, adr2 ∈ fs0, adr3 ∈ fs0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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